How to Find a Polariser Angle from Transmitted Intensity

A beam reaches a polarising filter at an intensity of \(24\ \mathrm{W\,m^{-2}}\), but only \(6\ \mathrm{W\,m^{-2}}\) comes out. What angle does the filter’s transmission axis make with the incoming light’s polarisation direction?

The smaller angle is \(60^\circ\). The filter transmits one-quarter of the incoming intensity, but that doesn’t mean its axis is one-quarter of \(90^\circ\). Intensity depends on the square of the cosine of the angle.

To find an unknown polariser angle, rearrange Malus’ law:

\[ I_{\text{out}}=I_{\text{in}}\cos^2\theta \]

Define the transmitted fraction as \(r\):

\[ r=\frac{I_{\text{out}}}{I_{\text{in}}} \]

Then:

\[ \theta=\cos^{-1}(\sqrt{r}) \]

Here, \(I_{\text{in}}\) is the intensity immediately before the filter, \(I_{\text{out}}\) is the intensity immediately after it, and \(\theta\) is the angle between the incoming electric-field direction and the filter’s transmission axis.

For our beam:

\[ \begin{aligned} r&=\frac{6}{24}=0.25\\ \theta&=\cos^{-1}(\sqrt{0.25})\\ &=60^\circ \end{aligned} \]

Use degree mode on your calculator. This gives the smaller angle between the axes, from \(0^\circ\) to \(90^\circ\). It does not necessarily tell you the filter’s absolute orientation.

01Why a second polariser makes the angle measurable

Imagine you’re trying to determine the orientation of a polarising filter using only a lamp and a light detector.

Your first apparatus has three parts, arranged in a straight line:

1. A lamp produces approximately unpolarised light.
2. The light passes through one rotatable polarising filter.
3. A photodetector measures the transmitted intensity.

Suppose the lamp produces an intensity of \(48\ \mathrm{W\,m^{-2}}\).

An ideal polariser transmits half the intensity of unpolarised light, so the detector reads \(24\ \mathrm{W\,m^{-2}}\).

Now rotate the filter through \(90^\circ\). Would the reading fall to zero?

No. It remains \(24\ \mathrm{W\,m^{-2}}\), assuming the source stays unpolarised and the polariser is ideal.

The incoming light has electric-field oscillations in many transverse directions. Rotating the filter changes which direction is transmitted, but not the average fraction of unpolarised light that passes through.

That means this original apparatus cannot determine the filter’s orientation from intensity alone.

The revision: add a fixed polariser

Keep the lamp, but place a fixed polariser before the rotating filter.

The light now travels through:

• The lamp, producing unpolarised light.
• The fixed polariser, which establishes a known electric-field direction.
• The rotating analyser, whose transmission axis can be adjusted.
• The photodetector, which measures the final intensity.

The first filter produces plane-polarised light. Its electric field oscillates along the first filter’s transmission axis.

The second filter is called the analyser because changing its orientation reveals information about the incoming polarisation.

What should the revised experiment show?

Keep the intensity entering the analyser constant at \(24\ \mathrm{W\,m^{-2}}\).

As the analyser rotates away from alignment, the transmitted intensity should fall. When its axis is perpendicular to the incoming polarisation, an ideal analyser transmits no light.

The following are illustrative theoretical readings, not measurements from a particular experiment.

Notice something interesting about the \(30^\circ\) and \(60^\circ\) readings. Their angles differ by only \(30^\circ\), yet their transmitted intensities differ by a factor of three.

The relationship isn’t linear.

Why does the cosine get squared?

Picture an electric-field arrow pointing along the first polariser’s transmission axis.

The analyser only transmits the component of that electric field parallel to its own transmission axis.

If the incident electric-field amplitude is \(E_{\text{in}}\), the transmitted amplitude is:

\[ E_{\text{out}}=E_{\text{in}}\cos\theta \]

The amplitude is the maximum magnitude of the oscillating electric field. Its units are volts per metre, \(\mathrm{V\,m^{-1}}\).

This is similar to finding the projection of an arrow onto a line. The line represents the analyser’s transmission axis.

But the analogy has a limit. A polariser doesn’t physically trim an arrow or slice off a geometric portion of a beam. It selectively transmits an electric-field component.

The important extra step is that electromagnetic-wave intensity is proportional to the square of electric-field amplitude:

\[ I\propto E^2 \]

Therefore:

\[ \frac{I_{\text{out}}}{I_{\text{in}}} =\cos^2\theta \]

That’s Malus’ law.

When you’re finding an unknown angle, you work backwards through these two operations: take the square root of the intensity ratio, then take the inverse cosine.

For a broader explanation of the forward calculation, see Malus’ Law for HSC Physics: Intensity and Polariser Angles.

02Work backwards from the right intensity

The biggest source of wrong answers isn’t usually the inverse cosine. It’s choosing the wrong initial intensity.

Suppose the beam in our experiment begins unpolarised with an intensity of \(48\ \mathrm{W\,m^{-2}}\).

It passes through two ideal filters:

The angle of the analyser is determined by:

\[ I_2=I_1\cos^2\theta \]

Not by:

\[ I_2=I_0\cos^2\theta \]

The analyser doesn’t receive the original \(48\ \mathrm{W\,m^{-2}}\). It receives \(24\ \mathrm{W\,m^{-2}}\).

Using \(6/48\) would incorrectly give an angle of approximately \(69.3^\circ\), instead of \(60^\circ\).

The half-intensity rule applies to ideal unpolarised light entering an ideal polariser. If the entering light is already plane-polarised, you cannot automatically halve it.

A reliable decision process is:

1. Identify the filter whose angle is unknown.
2. Find the intensity immediately before that filter and immediately after it.
3. Calculate \(r=I_{\text{out}}/I_{\text{in}}\).
4. Calculate \(\theta=\cos^{-1}(\sqrt r)\).
5. Interpret that angle relative to the incoming polarisation direction, not automatically relative to the horizontal.

For an ideal analyser, the ratio must satisfy \(0\leq r\leq1\). A value outside that range means the stated measurements or model are inconsistent.

03Why one intensity can correspond to two different orientations

A polariser’s transmission axis doesn’t have a forward and backward end. Turning a filter through \(180^\circ\) produces the same axis.

There’s another ambiguity.

Suppose the first polariser has a horizontal transmission axis at \(0^\circ\).

An analyser at \(60^\circ\) has the same transmitted intensity as one at \(120^\circ\).

Their smaller angular separations from the horizontal are both \(60^\circ\).

Mathematically:

\[ \cos^2 60^\circ=\cos^2 120^\circ \]

Both values equal \(0.25\).

A single reading cannot distinguish these orientations.

To tell which one is present, you need information beyond the original intensity. A controlled rotation provides that information.

04What the detector may be hiding

The ideal calculation assumes a fully plane-polarised beam entering the analyser, an ideal analyser, and reliable intensity measurements.

Real measurements can differ for several reasons.

A background reading is not transmitted light

A detector might produce a non-zero signal when the beam is blocked. Ambient light and electronic offsets are possible contributors.

If the detector reads 4 units with no light, those 4 units shouldn’t be counted as light transmitted by the analyser.

That’s why the previous example used \(48\) and \(12\), rather than the raw readings of \(52\) and \(16\).

Real polarisers might not extinguish the beam completely

An imperfect analyser can allow a small amount of light through even when its nominal transmission axis is perpendicular to the incident polarisation.

If the measured response still follows a cosine-squared dependence between its minimum and maximum readings, you can calibrate using those extremes.

Let:

• \(S\) be the detector reading at the unknown angle.
• \(S_{\parallel}\) be the reading with parallel axes.
• \(S_{\perp}\) be the reading with perpendicular axes.

For the calibrated cosine-squared model:

\[ r=\frac{S-S_{\perp}}{S_{\parallel}-S_{\perp}} \]

Then:

\[ \theta=\cos^{-1}(\sqrt r) \]

This is a correction for an apparatus whose response follows that model. It is not a replacement for the ideal physical law in every situation.

The source brightness, detector sensitivity, and other experimental conditions must remain consistent between readings.

What if the incoming light isn’t fully plane-polarised?

This is a different problem.

Malus’ law in its simple form assumes that the beam entering the analyser has a definite linear polarisation direction.

If that beam contains a mixture of different polarisation directions, one intensity measurement generally cannot be inverted to find a single meaningful angle.

You may need another polariser or additional measurements to determine what kind of light you’re dealing with.

A reading also becomes unreliable if the lamp changes brightness while you’re measuring. Measuring the intensity immediately before the analyser, or monitoring the source during the experiment, helps separate changing brightness from changing transmission.

The distinction is simple: an intensity ratio can determine an angle only when the physical assumptions that connect the two are justified.

05Questions and solutions

The questions below progress from a direct HSC calculation to harder inverse problems. All numerical readings are original, illustrative scenarios.

Unless stated otherwise, assume ideal polarisers, stable conditions during each individual measurement, and angles measured anticlockwise from the horizontal. Transmission axes are specified modulo \(180^\circ\).

Question 1

A plane-polarised beam has an intensity of \(36\ \mathrm{W\,m^{-2}}\) immediately before an analyser.

The transmitted intensity is \(9\ \mathrm{W\,m^{-2}}\).

Calculate the smaller angle between the incident polarisation direction and the analyser’s transmission axis.

Solution 1

The smaller angle is \(60^\circ\).

The transmitted intensity is one-quarter of the incident intensity:

\[ r=\frac{9}{36}=0.25 \]

Using inverse Malus’ law:

\[ \begin{aligned} \theta&=\cos^{-1}(\sqrt{0.25})\\ &=60^\circ \end{aligned} \]

The analyser transmits \(25\%\) of the incoming intensity.

A common mistake is to calculate \(\cos^{-1}(0.25)\) without taking the square root first. The cosine itself describes the electric-field amplitude ratio; its square describes the intensity ratio.

The result determines the smaller angular separation, not a unique absolute orientation.

Question 2

Unpolarised light with an initial intensity of \(120\ \mathrm{W\,m^{-2}}\) passes through three ideal polarisers.

Their transmission-axis orientations are:

• P1: \(15^\circ\)
• P2: unknown
• P3: \(105^\circ\)

Measurements show that the intensity after P2 is \(45\ \mathrm{W\,m^{-2}}\), and after P3 it is \(11.25\ \mathrm{W\,m^{-2}}\).

(a) Determine all possible orientations of P2 between \(0^\circ\) and \(180^\circ\).

(b) Explain why the reading after P3 does not identify which orientation is correct.

(c) P3 is then rotated to \(135^\circ\), with P1 and P2 unchanged. Calculate the two possible final intensities. Explain how this measurement would distinguish the orientations.

Solution 2

P2 can be oriented at \(45^\circ\) or \(165^\circ\). Both agree with the original measurements, but rotating P3 to \(135^\circ\) would produce either \(0\) or \(33.75\ \mathrm{W\,m^{-2}}\), allowing the two possibilities to be distinguished.

(a) Find the possible P2 orientations.

The first polariser transmits half the unpolarised intensity:

\[ I_1=\frac{120}{2} =60\ \mathrm{W\,m^{-2}} \]

P2 transmits \(45\ \mathrm{W\,m^{-2}}\).

\[ \begin{aligned} r&=\frac{45}{60}=0.75\\ \theta&=\cos^{-1}(\sqrt{0.75})\\ &=30^\circ \end{aligned} \]

The axis of P2 must be \(30^\circ\) from the axis of P1.

Therefore:

\[ 15^\circ+30^\circ=45^\circ \]

Or:

\[ 15^\circ-30^\circ=-15^\circ \]

Because \(-15^\circ\) is equivalent to \(165^\circ\), the two possible orientations are \(45^\circ\) and \(165^\circ\).

(b) Explain why the final reading is ambiguous.

If P2 is at \(45^\circ\), its separation from P3 at \(105^\circ\) is \(60^\circ\).

If P2 is at \(165^\circ\), its smaller separation from P3 is also \(60^\circ\).

Both possibilities give:

\[ \begin{aligned} I_3&=45\cos^2 60^\circ\\ &=45(0.25)\\ &=11.25\ \mathrm{W\,m^{-2}} \end{aligned} \]

This agrees with the measured intensity in both cases.

The tempting conclusion is that three filters and two measured intensities must be enough to reconstruct the middle orientation uniquely.

They aren’t. The two possible middle axes are symmetrically arranged around the final filter’s axis, so the last measurement provides no new distinction.

(c) Rotate P3 to \(135^\circ\).

If P2 is at \(45^\circ\), the new separation is:

\[ 135^\circ-45^\circ=90^\circ \]

The filters are crossed:

\[ I_3=45\cos^2 90^\circ =0\ \mathrm{W\,m^{-2}} \]

If P2 is at \(165^\circ\), the smaller separation is \(30^\circ\):

\[ \begin{aligned} I_3&=45\cos^2 30^\circ\\ &=45(0.75)\\ &=33.75\ \mathrm{W\,m^{-2}} \end{aligned} \]

A zero reading would identify P2 as \(45^\circ\). A reading of \(33.75\ \mathrm{W\,m^{-2}}\) would identify it as \(165^\circ\).

This is an example of designing another measurement to distinguish two physically possible explanations, rather than simply collecting more of the same data.

Question 3

A fixed polariser has its transmission axis at \(25^\circ\). A rotatable analyser follows it.

The source brightness changes between measurements, but the polarisation direction after the fixed filter remains constant.

The analyser begins at an unknown axis orientation \(\phi\). It is then rotated anticlockwise by \(30^\circ\) and again by another \(30^\circ\).

These measurements are recorded:

(a) Determine the initial orientation \(\phi\) between \(0^\circ\) and \(180^\circ\).

(b) If the intensity entering the analyser had remained \(48\ \mathrm{W\,m^{-2}}\) throughout, calculate the three transmitted intensities.

(c) Explain why comparing only the three recorded transmitted intensities could lead to an incorrect conclusion about the analyser’s orientation.

Solution 3

The analyser initially has an orientation of \(145^\circ\). If the incoming intensity had stayed constant at \(48\ \mathrm{W\,m^{-2}}\), the three output readings would have been \(12\), \(36\), and \(48\ \mathrm{W\,m^{-2}}\).

(a) Find the initial orientation.

First calculate the fraction transmitted at each setting.

At the initial setting:

\[ r_1=\frac{12}{48}=0.25 \]

At the second setting:

\[ r_2=\frac{54}{72}=0.75 \]

At the third setting:

\[ r_3=\frac{32}{32}=1 \]

These correspond to smaller axis separations of:

\[ \begin{aligned} \theta_1&=60^\circ\\ \theta_2&=30^\circ\\ \theta_3&=0^\circ \end{aligned} \]

The third reading is particularly useful. The analyser transmits all the incoming intensity, so its axis must be aligned with the fixed polariser at \(25^\circ\), modulo \(180^\circ\).

The analyser reached that orientation after rotating \(60^\circ\) anticlockwise.

Therefore:

\[ \begin{aligned} \phi+60^\circ&\equiv25^\circ \pmod{180^\circ}\\ \phi&\equiv-35^\circ\\ &\equiv145^\circ \end{aligned} \]

So the initial analyser orientation was \(145^\circ\).

Check the intermediate setting:

\[ 145^\circ+30^\circ=175^\circ \]

The difference between \(175^\circ\) and \(25^\circ\) is \(150^\circ\), giving a smaller axis separation of \(30^\circ\).

Its predicted transmission fraction is:

\[ \cos^2 30^\circ=0.75 \]

That matches the second measurement.

(b) Keep the incoming intensity constant.

Use \(I_{\text{in}}=48\ \mathrm{W\,m^{-2}}\) at all three settings.

Initially:

\[ I_1=48\cos^2 60^\circ =12\ \mathrm{W\,m^{-2}} \]

After rotating \(30^\circ\):

\[ I_2=48\cos^2 30^\circ =36\ \mathrm{W\,m^{-2}} \]

After rotating another \(30^\circ\):

\[ I_3=48\cos^2 0^\circ =48\ \mathrm{W\,m^{-2}} \]

(c) Why the raw readings mislead.

The recorded transmitted intensities were \(12\), \(54\), and \(32\ \mathrm{W\,m^{-2}}\).

Someone looking only at these values might conclude that the analyser became better aligned at the second setting, then less aligned at the third.

But the intensity entering the analyser changed from \(72\) to \(32\ \mathrm{W\,m^{-2}}\) between the second and third measurements.

The third setting actually has perfect alignment. The lower transmitted intensity comes entirely from the lower incoming intensity.

The important measurement is the transmitted fraction, not the transmitted intensity in isolation.

Question 4

A laboratory uses a polariser and an imperfect analyser to measure an unknown angle.

The incident light is fully plane-polarised and the source is initially steady.

The photodetector produces a constant dark reading even without light. The analyser also allows a small amount of optical leakage when the axes are crossed.

The laboratory assumes the detector response follows this calibrated model:

\[ S=S_\perp+(S_\parallel-S_\perp)\cos^2\theta \]

Here, \(S\) is the raw detector reading at angle \(\theta\), while \(S_\perp\) and \(S_\parallel\) are the readings for crossed and parallel axes at the same source brightness.

The following illustrative measurements are obtained:

(a) Calculate the smaller angle between the axes.

(b) Explain why subtracting only the dark reading of 6 units gives an incorrect angle.

(c) The source brightness is doubled without moving the analyser. Assume optical signals scale directly with source brightness while the dark reading stays constant. Predict the new crossed, parallel, and unknown-angle readings, and verify that the corrected angle remains unchanged.

Solution 4

The angle is approximately \(63.4^\circ\). After doubling the source brightness, the crossed, parallel, and unknown-angle readings become 14, 134, and 38 units, respectively.

(a) Use the measured minimum and maximum.

From the calibrated model:

\[ \cos^2\theta =\frac{S-S_\perp}{S_\parallel-S_\perp} \]

Substitute the readings:

\[ \begin{aligned} r&=\frac{22-10}{70-10}\\ &=\frac{12}{60}\\ &=0.20 \end{aligned} \]

Therefore:

\[ \begin{aligned} \theta&=\cos^{-1}(\sqrt{0.20})\\ &\approx63.4^\circ \end{aligned} \]

(b) Explain why the dark reading alone is insufficient.

The crossed reading is 10 units, but the lamp-off dark reading is only 6 units.

The difference is:

\[ 10-6=4\text{ units} \]

Those 4 units are optical leakage under the stated model.

If you subtract only the dark reading, you obtain:

\[ \frac{22-6}{70-6} =\frac{16}{64}=0.25 \]

That would incorrectly suggest \(60^\circ\).

The mistake is treating all light reaching the detector as if it followed the ideal analyser’s simple cosine-squared transmission with zero optical signal at crossed axes.

The calibrated model explicitly includes a non-zero minimum optical transmission.

(c) Double the source brightness.

The dark contribution stays at 6 units. Each light-dependent contribution doubles.

For the crossed axes, the original light-dependent signal is \(10-6=4\) units.

\[ S_\perp'=6+2(4)=14 \]

For parallel axes:

\[ \begin{aligned} S_\parallel'&=6+2(70-6)\\ &=134 \end{aligned} \]

For the unknown angle:

\[ \begin{aligned} S'&=6+2(22-6)\\ &=38 \end{aligned} \]

All three readings are in the same arbitrary detector units.

Now calculate the calibrated fraction again:

\[ \begin{aligned} r'&=\frac{38-14}{134-14}\\ &=\frac{24}{120}\\ &=0.20 \end{aligned} \]

The inferred angle remains:

\[ \theta=\cos^{-1}(\sqrt{0.20}) \approx63.4^\circ \]

The raw detector reading changed, but the analyser’s orientation didn’t.

This result depends on the stated response model and stable calibration. If the detector response were nonlinear or the leakage changed unpredictably, the calculation would need a more suitable model.

Question 5

A sealed light source illuminates two ideal polarisers in sequence.

The first polariser, P, can rotate. The second, A, is an independently rotatable analyser. A photodetector measures the transmitted intensity after A.

The source is steady, but its polarisation state is unknown.

You are told that the source is either:

• Completely unpolarised light of unknown intensity, or
• Completely plane-polarised light of unknown intensity and polarisation direction.

No intensity measurement is available before P.

Five experiments are performed. For each trial, the filters are set to the orientations shown, and the transmitted intensity is recorded.

(a) Using only Trials 1-4, determine all source models consistent with the data. State their incident intensity and, where relevant, possible polarisation directions.

(b) Use Trial 5 to identify the actual source model and its polarisation direction. Explain why the larger transmitted intensity does not violate energy conservation.

(c) A student suggests that measuring the intensity at every analyser angle, while keeping P fixed at \(0^\circ\), would have revealed the source’s original polarisation state without Trial 5. Is the student correct? Justify your answer.

Solution 5

The source is completely plane-polarised, with incident intensity \(60\ \mathrm{W\,m^{-2}}\) and polarisation direction \(45^\circ\).

The surprising part is that Trials 1-4 cannot distinguish this source from unpolarised light. Even the measurement using a first polariser at \(90^\circ\) leaves both explanations possible.

(a) Find the models that agree with Trials 1-4.

First consider an unpolarised source.

In Trial 1, P and A are aligned at \(0^\circ\).

The first ideal polariser transmits half the unpolarised intensity, and the aligned analyser transmits all of that remaining light.

Let \(U\) represent the incident intensity of the unpolarised source.

\[ \frac{U}{2}=30 \]

Therefore:

\[ U=60\ \mathrm{W\,m^{-2}} \]

This model predicts Trial 2 correctly:

\[ \begin{aligned} I_2&=30\cos^2 60^\circ\\ &=7.5\ \mathrm{W\,m^{-2}} \end{aligned} \]

It also predicts zero for the crossed filters in Trial 3.

In Trial 4, both filters are rotated to \(90^\circ\). The source is still unpolarised, so P still transmits half its intensity.

The detector again reads:

\[ I_4=\frac{60}{2} =30\ \mathrm{W\,m^{-2}} \]

Thus all four trials are consistent with unpolarised incident light of intensity \(60\ \mathrm{W\,m^{-2}}\).

Now consider the second model: a completely plane-polarised source.

Let its incident intensity be \(T\), and let its polarisation direction be \(\alpha\).

In Trial 1, the first polariser’s axis is \(0^\circ\), and the analyser is aligned with it.

The detector therefore measures:

\[ T\cos^2\alpha=30 \]

In Trial 4, P is at \(90^\circ\), and A is aligned with P.

The incident polarisation is now projected onto the perpendicular direction:

\[ T\sin^2\alpha=30 \]

Add these two equations:

\[ T(\cos^2\alpha+\sin^2\alpha)=60 \]

Using the identity:

\[ \cos^2\alpha+\sin^2\alpha=1 \]

We obtain:

\[ T=60\ \mathrm{W\,m^{-2}} \]

Substitute this into the first equation:

\[ \begin{aligned} 60\cos^2\alpha&=30\\ \cos^2\alpha&=0.5 \end{aligned} \]

The possible incident polarisation directions are:

\[ \alpha=45^\circ \quad\text{or}\quad135^\circ \]

Both produce \(30\ \mathrm{W\,m^{-2}}\) after P at \(0^\circ\), and also after P at \(90^\circ\).

Trials 2 and 3 do not resolve this ambiguity. Once P is fixed at \(0^\circ\), both possible plane-polarised sources produce an outgoing beam of intensity \(30\ \mathrm{W\,m^{-2}}\), polarised along \(0^\circ\).

The analyser must therefore give \(7.5\ \mathrm{W\,m^{-2}}\) at \(60^\circ\) and zero at \(90^\circ\), exactly as measured.

So Trials 1-4 permit three source possibilities:

• Unpolarised light, intensity \(60\ \mathrm{W\,m^{-2}}\).
• Plane-polarised light, intensity \(60\ \mathrm{W\,m^{-2}}\), direction \(45^\circ\).
• Plane-polarised light, intensity \(60\ \mathrm{W\,m^{-2}}\), direction \(135^\circ\).

The measurements don’t yet distinguish them.

(b) Use Trial 5 to identify the source.

In Trial 5, both P and A are set to \(45^\circ\).

The detector reads \(60\ \mathrm{W\,m^{-2}}\).

Consider the unpolarised model first.

The first filter would still transmit only half the incident intensity:

\[ I_P=\frac{60}{2} =30\ \mathrm{W\,m^{-2}} \]

Since A is aligned with P, the final reading would also be \(30\ \mathrm{W\,m^{-2}}\).

That contradicts the observed \(60\ \mathrm{W\,m^{-2}}\).

So the unpolarised model is impossible.

Now consider the plane-polarised source at \(135^\circ\).

Its polarisation direction is perpendicular to P at \(45^\circ\).

\[ \begin{aligned} I_P&=60\cos^2 90^\circ\\ &=0\ \mathrm{W\,m^{-2}} \end{aligned} \]

This also contradicts Trial 5.

Finally, consider the source polarised at \(45^\circ\).

Its electric field is aligned with P, so the first ideal polariser transmits the entire incident intensity:

\[ \begin{aligned} I_P&=60\cos^2 0^\circ\\ &=60\ \mathrm{W\,m^{-2}} \end{aligned} \]

A is also at \(45^\circ\), so it transmits all the remaining light.

The final reading is:

\[ I_5=60\ \mathrm{W\,m^{-2}} \]

This matches Trial 5 exactly.

Therefore, the source is fully plane-polarised at \(45^\circ\), with incident intensity \(60\ \mathrm{W\,m^{-2}}\).

There is no violation of energy conservation. The transmitted intensity increased compared with Trial 1 because the first polariser was rotated into alignment with the original electric-field direction.

The filters did not create extra light. They simply discarded less of the incident intensity.

(c) Would rotating only the analyser identify the original source?

No. Rotating A through every angle while P remains fixed at \(0^\circ\) cannot distinguish these source models.

In all three models consistent with Trials 1-4, the beam leaving P at \(0^\circ\) has the same properties:

• Intensity \(30\ \mathrm{W\,m^{-2}}\).
• Plane-polarisation direction \(0^\circ\).

If A is at angle \(\phi\), every model therefore predicts:

\[ I_A=30\cos^2\phi \]

A complete scan from \(0^\circ\) to \(180^\circ\) would produce the same intensity curve for all three possible sources.

The detector is observing light after P has selected one polarisation component. Information about the original beam’s polarisation state has been lost in that first filtering step.

That is the central experimental limitation.

Rotating A tells you about the light arriving at A. Rotating P, while keeping A aligned with it, tests how the original source transmits through different polarisation directions.

Trial 5 changes the first filter’s orientation and therefore reveals information that analyser-only measurements cannot recover.

This is why finding a polariser angle is not always just an inverse-cosine calculation. You first need to know which light intensity is entering the filter, what polarisation state that light has, and whether your measurements distinguish all physically possible orientations.

The next useful challenge is reconstructing the behaviour of a longer optical system, where every filter changes both the intensity and the polarisation direction of the beam that reaches the next one. The same reasoning carries into How to Apply Malus’ Law to Three Polarising Filters.